Difference between revisions of "2016 AIME II Problems/Problem 10"
(→Solution) |
(→Solution) |
||
Line 1: | Line 1: | ||
Triangle <math>ABC</math> is inscribed in circle <math>\omega</math>. Points <math>P</math> and <math>Q</math> are on side <math>\overline{AB}</math> with <math>AP<AQ</math>. Rays <math>CP</math> and <math>CQ</math> meet <math>\omega</math> again at <math>S</math> and <math>T</math> (other than <math>C</math>), respectively. If <math>AP=4,PQ=3,QB=6,BT=5,</math> and <math>AS=7</math>, then <math>ST=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Triangle <math>ABC</math> is inscribed in circle <math>\omega</math>. Points <math>P</math> and <math>Q</math> are on side <math>\overline{AB}</math> with <math>AP<AQ</math>. Rays <math>CP</math> and <math>CQ</math> meet <math>\omega</math> again at <math>S</math> and <math>T</math> (other than <math>C</math>), respectively. If <math>AP=4,PQ=3,QB=6,BT=5,</math> and <math>AS=7</math>, then <math>ST=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
import cse5; | import cse5; | ||
Line 35: | Line 35: | ||
Now Law of Sines on <math>\triangle ACS</math>, <math>\triangle SCT</math>, and <math>\triangle TCB</math> yields <cmath>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.</cmath>Hence <cmath>\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},</cmath>so <cmath>TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.</cmath>Hence <math>ST=\tfrac{35}8</math> and the requested answer is <math>35+8=\boxed{43}</math>. | Now Law of Sines on <math>\triangle ACS</math>, <math>\triangle SCT</math>, and <math>\triangle TCB</math> yields <cmath>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.</cmath>Hence <cmath>\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},</cmath>so <cmath>TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.</cmath>Hence <math>ST=\tfrac{35}8</math> and the requested answer is <math>35+8=\boxed{43}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}</math> |
Revision as of 22:24, 17 March 2016
Triangle is inscribed in circle
. Points
and
are on side
with
. Rays
and
meet
again at
and
(other than
), respectively. If
and
, then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let
,
, and
. Note that since
we have
, so by the Ratio Lemma
Similarly, we can deduce
and hence
.
Now Law of Sines on ,
, and
yields
Hence
so
Hence
and the requested answer is
.
Solution 2
Projecting through we have
which easily gives